Falling Rock on Planet X. Solving for 'g'.

[WCSpeidel]

Homework Statement

After performing a quadratic curve fit tool in Pro Logger, I was given the values: A = -1.426E-05 +/- 2.386E-06 , B = 0.02193 +/- 0.001581, C = 0.4785 +/- 0.1905.

With this information, how is “g” for Planet X determined from the parameters of the curve fit equation? And what is the value of “g” that you determined for Planet X?

Homework Equations

y = At^2+Bt+C or y = at^2/2 + v0t + y0

y = 1/2gt^2 or g =2y/t^2

The Attempt at a Solution

My first thought is that since there was no initial velocity and no initial position given, I am focusing in on y = at^2/2 or the y = 1/2gt^2.

Since the seconds equation has the most obvious means to find "g", that is what I have tried with no success.

Example: 168 = 1/2(g)(4.00)^2 --> I am resulting with a 'g' value of 21

While for: 107 = 1/2(g)(3.00)2 --> I come up with a 'g' value of 23.8

Am I using an incorrect method of finding the value of "g"? Any help would be most appreciated.

 

[mfb]

Am I using an incorrect method of finding the value of "g"?

Yes. There is no given initial velocity or position, so you cannot assume that they would be zero (in fact, they are not). You'll need a more general formula for your physical motion, just y = 1/2gt² is not enough.

If you have that, you can just compare coefficients, it is not necessary to plug in values for specific t.

 

[td21]

I believe this is free fall from the numbers given. The number on the L.H.S. is the distance from the original position.
With no initial velocity, you method is correct.
The value of g is in general different for different calculations due to experimental errors, what you need to do is to find the mean or average of them. I suggest that this can be done by finding the slope of y vs t^2 graph and the value of g can be found by multiplying the slope by a number(which is?).

 

[WCSpeidel]

Yes. There is no given initial velocity or position, so you cannot assume that they would be zero (in fact, they are not). You'll need a more general formula for your physical motion, just y = 1/2gt² is not enough.

If you have that, you can just compare coefficients, it is not necessary to plug in values for specific t.

And I can understand that considering I am coming up with values that do not make sense. With that said, would the theoretical formula y = At^2/2 + Bt + C be a more accurate formula?

Even with this formula, understanding that the variables are ‘y’ as position in meters and ‘t’ as the time elapsed in seconds. And using the coefficients ‘A’ as acceleration, ‘B’ as initial velocity and ‘C’ as the initial position. I am still finding myself stuck when trying to achieve of value for 'g'.

 

[WCSpeidel]

The value of g is in general different for different calculations due to experimental errors, what you need to do is to find the mean or average of them.

After taking the nine different measurement values of the position and time, the average of the nine comes out to be 24.12 m/s. The 'g' values range from 53.6 m/s in the beginning measurements down to 16.5 m/s for the last measurement.

My gut reaction is that there should just be one 'g' value for the entire experiment considering the questions asks, "what is the value of “g” that you determined for Planet X?"

 

[td21]

Have you square the t? If so that may be the one. If you have not submitted the hw, i suggest you plot the y vs tsquare graph and use the best fitting line in excel to connect the data point. Then next is to find the slope of that best fitting line and multiply the slope by 2.